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Update: A year ago, but the first answer is not clear with me. I bounty this question again.

My question: I am looking for a proof or counterexample to the following inequality:

If $n \in \mathbb{N}$, and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \cdots \ge \alpha_n \ge 0$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \ge \prod_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

If $n \in \mathbb{N}$ and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \cdots \le \alpha_n$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \le \prod_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

Moreover equality holds in either case if and only if $a_1=a_2=a_3=...=a_n$.

By computation the inequality is true when $n=2$ and $n=3$.

For instance, if $n=2$ then for the first inequality, we have

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \ge (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$

since the difference is $a_1^{\alpha_1}a_2^{\alpha_2} - a_1^{\alpha_2}a_2^{\alpha_1}$, which is positive by a rearrangement inequality.

Update: A year ago, but the first answer is not clear with me. I bounty this question again.

My question: I am looking for a proof or counterexample to the following inequality:

If $n \in \mathbb{N}$, and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \cdots \ge \alpha_n \ge 0$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \ge \prod_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

If $n \in \mathbb{N}$ and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \cdots \le \alpha_n$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \le \prod_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

Moreover equality holds in either case if and only if $a_1=a_2=a_3=...=a_n$.

[The original question claimed a proof when $n=2$ or $n=3$.]

Update: A year ago, but the first answer is not clear with me. I bounty this question again.

My question: I am looking for a proof of the inequality as follows, is the inequality correct?

Inequality: If $n$ be positive integer $n \ge 2$ and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \cdots \ge \alpha_n \ge 0$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \ge \Pi_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

If $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \cdots \le \alpha_n$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \le \Pi_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

Equality holdes if only if $a_1=a_2=a_3=...=a_n$

By my computation the inequality as follows good be true for case $n=2$ and $n=3$. So I conjecture the inequality above true with $n \ge 2$.

Case n=2:

If $a_1 \ge a_2 \ge 0$ and $\alpha_1 \ge \alpha_2 \ge 0$ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \ge (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$

and

If $a_1 \ge a_2 \ge 0$ and $0 \le \alpha_1 \le \alpha_2 $ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \le (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$

Case n=3:

If $a_1 \ge a_2 \ge a_3 \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \alpha_3 \ge 0$ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2}+a_3^{\alpha_3})^3 \ge (a_1^{\alpha_1}+a_1^{\alpha_2}+a_1^{\alpha_3})(a_2^{\alpha_1}+a_2^{\alpha_2}+a_2^{\alpha_3})(a_3^{\alpha_1}+a_3^{\alpha_2}+a_3^{\alpha_3})$$

and

If $a_1 \ge a_2 \ge a_3 \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \alpha_3 $ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2}+a_3^{\alpha_3})^3 \le (a_1^{\alpha_1}+a_1^{\alpha_2}+a_1^{\alpha_3})(a_2^{\alpha_1}+a_2^{\alpha_2}+a_2^{\alpha_3})(a_3^{\alpha_1}+a_3^{\alpha_2}+a_3^{\alpha_3})$$

Update: A year ago, but the first answer is not clear with me. I bounty this question again.

My question: I am looking for a proof or counterexample to the following inequality:

If $n \in \mathbb{N}$, and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \cdots \ge \alpha_n \ge 0$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \ge \prod_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

If $n \in \mathbb{N}$ and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \cdots \le \alpha_n$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \le \prod_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

Moreover equality holds in either case if and only if $a_1=a_2=a_3=...=a_n$.

By computation the inequality is true when $n=2$ and $n=3$.

For instance, if $n=2$ then for the first inequality, we have

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \ge (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$

since the difference is $a_1^{\alpha_1}a_2^{\alpha_2} - a_1^{\alpha_2}a_2^{\alpha_1}$, which is positive by a rearrangement inequality.

Update: The first answer is not clear with me. I bounty this question again.

My question: I am looking for a proof of the inequality as follows, is the inequality correct?

Inequality: If $n$ be positive integer $n \ge 2$ and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \cdots \ge \alpha_n \ge 0$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \ge \Pi_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

If $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \cdots \le \alpha_n$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \le \Pi_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

Equality holdes if only if $a_1=a_2=a_3=...=a_n$

By my computation the inequality as follows good be true for case $n=2$ and $n=3$. So I conjecture the inequality above true with $n \ge 2$.

Case n=2:

If $a_1 \ge a_2 \ge 0$ and $\alpha_1 \ge \alpha_2 \ge 0$ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \ge (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$

and

If $a_1 \ge a_2 \ge 0$ and $0 \le \alpha_1 \le \alpha_2 $ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \le (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$

Case n=3:

If $a_1 \ge a_2 \ge a_3 \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \alpha_3 \ge 0$ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2}+a_3^{\alpha_3})^3 \ge (a_1^{\alpha_1}+a_1^{\alpha_2}+a_1^{\alpha_3})(a_2^{\alpha_1}+a_2^{\alpha_2}+a_2^{\alpha_3})(a_3^{\alpha_1}+a_3^{\alpha_2}+a_3^{\alpha_3})$$

and

If $a_1 \ge a_2 \ge a_3 \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \alpha_3 $ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2}+a_3^{\alpha_3})^3 \le (a_1^{\alpha_1}+a_1^{\alpha_2}+a_1^{\alpha_3})(a_2^{\alpha_1}+a_2^{\alpha_2}+a_2^{\alpha_3})(a_3^{\alpha_1}+a_3^{\alpha_2}+a_3^{\alpha_3})$$

Update: A year ago, but the first answer is not clear with me. I bounty this question again.

My question: I am looking for a proof of the inequality as follows, is the inequality correct?

Inequality: If $n$ be positive integer $n \ge 2$ and $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \cdots \ge \alpha_n \ge 0$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \ge \Pi_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

If $a_1 \ge a_2 \ge \cdots \ge a_n \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \cdots \le \alpha_n$ then

$${\left(\sum_{i=1}^{n}{a_i^{\alpha_i}} \right)}^n \le \Pi_{i=1}^{n}{\left(\sum_{j=1}^{n}{a_i^{\alpha_j}} \right)}$$

Equality holdes if only if $a_1=a_2=a_3=...=a_n$

By my computation the inequality as follows good be true for case $n=2$ and $n=3$. So I conjecture the inequality above true with $n \ge 2$.

Case n=2:

If $a_1 \ge a_2 \ge 0$ and $\alpha_1 \ge \alpha_2 \ge 0$ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \ge (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$

and

If $a_1 \ge a_2 \ge 0$ and $0 \le \alpha_1 \le \alpha_2 $ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2})^2 \le (a_1^{\alpha_1}+a_1^{\alpha_2})(a_2^{\alpha_1}+a_2^{\alpha_2})$$

Case n=3:

If $a_1 \ge a_2 \ge a_3 \ge 0$ and $\alpha_1 \ge \alpha_2 \ge \alpha_3 \ge 0$ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2}+a_3^{\alpha_3})^3 \ge (a_1^{\alpha_1}+a_1^{\alpha_2}+a_1^{\alpha_3})(a_2^{\alpha_1}+a_2^{\alpha_2}+a_2^{\alpha_3})(a_3^{\alpha_1}+a_3^{\alpha_2}+a_3^{\alpha_3})$$

and

If $a_1 \ge a_2 \ge a_3 \ge 0$ and $0 \le \alpha_1 \le \alpha_2 \le \alpha_3 $ then

$$(a_1^{\alpha_1}+a_2^{\alpha_2}+a_3^{\alpha_3})^3 \le (a_1^{\alpha_1}+a_1^{\alpha_2}+a_1^{\alpha_3})(a_2^{\alpha_1}+a_2^{\alpha_2}+a_2^{\alpha_3})(a_3^{\alpha_1}+a_3^{\alpha_2}+a_3^{\alpha_3})$$